Physics, asked by diptijaveri487, 1 year ago

A car starts from rest and moves with uniform acceleration 'a' on a straight road from time t=0 to t=T. After that, a constant deceleration brings it to rest. In this process the average speed of the car is - (a) aT/4 (b) 3aT/2 (c) aT/2 (d) aT

Answers

Answered by SahalPathan
28

Answer:

aT/2

Explanation:

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Answered by sourasghotekar123
1

Answer:

From the given options, the correct option is C)  a T/2

Explanation:

V = a T/2

Explanation: We are aware that

total distance travelled / total time taken = average speed

The car begins at rest and then accelerates with an acceleration 'a' after a time 'T' to a maximum velocity 'v'. Then it decelerates with the same acceleration and comes to a complete stop in time 'T' more. Let the distance travelled by it in the first case be s and the distance travelled in the second case be s'.

so, average speed will be = (s + s') / (T+T)

We already know that v2 - u2 = 2as, so

s = v2/2a

similarly

s' = v2/2a

We get from v = u + a (for the first case)

v = aT (t = T)

As a result of using this value of v in the preceding two equations, we obtain

2a2T2/a = 2aT2 s = s'

the average velocity is now

V = (s + s')/2T = (aT2 + aT2)/2T = (aT2 + aT2) / 2T

as a result,

V = aT/2

#SPJ6

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